From the following combinations of physical constants (expressed...

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:

Options

A $\frac{\mathrm{ch}}{2 \pi \varepsilon_{\mathrm{o}}^2}$
B $\frac{\mathrm{e}^2}{2 \pi \varepsilon_{\mathrm{o}} \mathrm{Gm}_{\mathrm{e}}^2}$
C $\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{\mathrm{c}^2} \frac{\mathrm{G}}{\mathrm{he}^2}$
D $\frac{2 \pi \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}{\mathrm{ce}^2} \frac{\mathrm{h}}{\mathrm{G}}$

Correct answer

B. $\frac{\mathrm{e}^2}{2 \pi \varepsilon_{\mathrm{o}} \mathrm{Gm}_{\mathrm{e}}^2}$

Step-by-step solution

The dimensional formulae of

\begin{aligned}&\mathrm{e}=\left[\mathrm{M}^0 \mathrm{ L}^0 \mathrm{ T}^1 \mathrm{ A}^1\right] \\ &\varepsilon_0=\left[\mathrm{M}^{-1} \mathrm{ L}^3 \mathrm{ T}^4 \mathrm{ A}^2\right] \\ &\mathrm{G}= \left[\mathrm{M}^{-1} \mathrm{ L}^3 \mathrm{ T}^{-2}\right]\\ &\text{ and } \mathrm{m}_{\mathrm{e}}= \left[\mathrm{M}^1 \mathrm{ L}^0 \mathrm{ T}^0\right]\\ &\text{ Now, } \frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}^2}\\ &=\frac{\left[\mathrm{M}^0 \mathrm{ L}^0 \mathrm{ T}^1 \mathrm{ A}^1\right]^2}{2 \pi\left[\mathrm{M}^{-1} \mathrm{ L}^{-3} \mathrm{ T}^4 \mathrm{ A}^2\right]\left[\mathrm{M}^{-1} \mathrm{ L}^3 \mathrm{ T}^{-2}\right]\left[\mathrm{M}^1 \mathrm{ L}^0 \mathrm{ T}^0\right]^2} \end{aligned}

\begin{aligned}&=\frac{\left[\mathrm{T}^2 \mathrm{ A}^2\right]}{2 \pi\left[\mathrm{M}^{-1-1+2} \mathrm{ L}^{-3+3} \mathrm{ T}^{4-2} \mathrm{ A}^2\right]}\\ &=\frac{\left[\mathrm{T}^2 \mathrm{ A}^2\right]}{2 \pi\left[\mathrm{M}^0 \mathrm{ L}^0 \mathrm{ T}^2 \mathrm{ A}^2\right]}=\frac{1}{2 \pi} \end{aligned}

\because \frac{1}{2 \pi}

is dimensionless thus the combination

\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}^2}$ would have the same value in different systems of units.

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